# 6.5: The temperature (heat) equation

We now resume our quest for a closed set of equations to describe the flow of a Newtonian fluid. We previously assumed mass and momentum conservation, resulting in the density equation and the Navier-Stokes momentum equation Equation 6.3.37. This collection of equations totals 4, but involves 5 unknowns: $$ρ$$, $$p$$, and the three components of $$\vec{u}$$.

We have now invoked a new assumption, namely energy conservation in the form of the 1st law of thermodynamics. This will allow us to add a new equation to the set. We first convert Equation 6.4.29 to Eulerian form. Two terms require conversion to volume integrals.

• We apply Cauchy’s lemma to the left -hand side, resulting in:

$\frac{D}{D t} \int_{V_{m}} \rho \mathscr{I} d V=\int_{V_{m}} \rho \frac{D \mathscr{I}}{D t} d V.$

• The heat loss can be converted using the divergence theorem (section 4.2.3):

$\oint_{A_{m}} \vec{q} \cdot \hat{n} d A=\int_{V_{m}} \vec{\nabla} \cdot \vec{q} d V$

We now have

$\int_{V_{m}}\left(\rho \frac{D \mathscr{I}}{D t}+\vec{\nabla} \cdot \vec{q}+p \vec{\nabla} \cdot \vec{u}-\rho \varepsilon\right) d V=0.\qquad .$

We conclude as usual that the integrand must be zero everywhere, resulting in:

$\rho \frac{D \mathscr{I}}{D t}=k \nabla^{2} T-\vec{\nabla} \cdot \vec{q}_{r a d}-p \vec{\nabla} \cdot \vec{u}+\rho \varepsilon,\label{eqn:1}$

where Equation 6.4.27 has been used for the heat flux. We have gained a new equation, but have also introduced two new unknowns, the internal energy I and the temperature $$T$$. (Note that neither $$\varepsilon$$ nor $$\vec{q}_{rad}$$ counts as an unknown. The former is determined by the velocity field via the strain, $$\varepsilon = 2v e^2_{ij}$$, while we assume that the latter is specified independently.) This leaves us short two pieces of information.

## 6.5.1 Specific heat capacity

Our next goal is to eliminate internal energy from the problem by establishing a relationship between it and temperature. We will consider two possible approaches, each based on an assumption about the nature of the temperature changes that can be illustrated with a simple lab experiment.

#### Incompressible heating

Suppose that a fluid sample is contained in a closed, rigid vessel wherein it is slowly heated. We keep track of the heat input and the resulting temperature rise, and the two turn out to be approximately proportional:

$\left(\frac{\partial Q}{\partial T}\right)_{\upsilon}=C_{\upsilon}.\label{eqn:2}$

Here $$Q$$ is the specific heat content, i.e., heat content per unit mass, typically measured in joules per kilogram. The subscript $$\upsilon$$ on the partial derivative specifies that the specific volume $$\upsilon=\rho-1$$ is held fixed while $$Q$$ and $$T$$ are changing. $$C_\upsilon$$ is called the specific heat capacity at constant volume, and can be regarded as constant if the range of temperatures is not too wide. Typical values are

$c_{\upsilon}=\left\{\begin{array}{l} 42000 k_{k}-1_{k}-1, \text { in\ water } \\ 7000 k g^{-1} K^{-1}, \text {in\ air. } \end{array}\right.\label{eqn:3}$

Now because this incompressible fluid does not expand or contract, changes in internal energy are due entirely to changes in heat content, so

$\frac{D \mathscr{I}}{D t}=\frac{D Q}{D t}=C_{\upsilon} \frac{D T}{D t}.$

We can now rewrite Equation $$\ref{eqn:1}$$ as

$\rho C_{v} \frac{D T}{D t}=k \nabla^{2} T-\vec{\nabla} \cdot \vec{q}_{r a d}+\rho \varepsilon,\label{eqn:4}$

keeping in mind that $$\vec{\nabla}\cdot\vec{u}=0$$. We have now succeeded in eliminating $$E$$, but the solution only works if $$\vec{\nabla}\cdot\vec{u}=0$$. Compressibility effects are not always negligible, especially in gases. To allow for that possibility, we imagine a slightly different experiment.

Isobaric heating

Suppose that we once again apply heat to a sample of fluid, but that the fluid is enclosed not in a rigid container but rather in a flexible membrane, like a balloon. As a result, the fluid can expand or contract freely, but the pressure does not change. (We assume that our balloon does not change altitude significantly, as would a weather balloon.) The process is therefore designated as isobaric.

For this process we define a new thermodynamic variable called the specific enthalpy, $$H$$. When a system changes slowly, the change in enthalpy is given by ∆H = ∆I +∆(pυ). In an isobaric process, this becomes $$\Delta H = \Delta \mathscr{I} + p\Delta \upsilon$$. For a given change in temperature, the change in enthalpy is given by

$\left(\frac{\partial H}{\partial T}\right)_{p}=C_{p}.\label{eqn:5$

$$C_p$$ is called the specific heat capacity at constant pressure, and is approximately constant for small temperature changes. Typical values are

$C_{p}=\left\{\begin{array}{l} 4200 \mathrm{J} \mathrm{kg}^{-1} \mathrm{K}^{-1}, \text {in water } \\ 1000 \mathrm{J} \mathrm{kg}^{-1} \mathrm{K}^{-1}, \text {in air. } \end{array}\right.\label{eqn:6}$

For a fluid parcel undergoing this isobaric change, the material derivative of the enthalpy is

$\frac{D H}{D t}=\frac{D \mathscr{I}}{D t}+p \frac{D \upsilon}{D t}=C_{p} \frac{D T}{D t}.\label{eqn:7}$

The material derivative of $$\upsilon$$ can be written as

$\frac{D \upsilon}{D t}=\frac{D \rho^{-1}}{D t}=-\rho^{-2} \frac{D \rho}{D t}=-\rho^{-2}(-\rho \vec{\nabla} \cdot \vec{u})=\rho^{-1} \vec{\nabla} \cdot \vec{u},\label{eqn:8}$

where mass conservation Equation 6.2.5 has been invoked. Multiplying Equation $$\ref{eqn:7}$$ by density and substituting Equation $$\ref{eqn:8}$$, we have

$\rho \frac{D H}{D t}=\rho \frac{D \mathscr{I}}{D t}+p \vec{\nabla} \cdot \vec{u}=\rho C_{p} \frac{D T}{D t}$

which, together with Equation $$\ref{eqn:1}$$, gives

$\rho C_{p} \frac{D T}{D t}=k \nabla^{2} T-\vec{\nabla} \cdot \vec{q}_{r a d}+\rho \varepsilon.\label{eqn:9}$

## 6.5.2 The heat equation

The two temperature equations that hold in the incompressible and isobaric approximations, Equation $$\ref{eqn:4}$$ and $$\ref{eqn:9}$$, differ only in the choice of the specific heat capacity:

$\rho\left(C_{\upsilon} \text { or } C_{p}\right) \frac{D T}{D t}=k \nabla^{2} T-\vec{\nabla} \cdot \vec{q}_{r a d}+\rho \varepsilon.$

Which approximation is better? In water, there is virtually no difference, because $$C_\upsilon$$ is nearly equal to $$C_p$$. In air, compressibility can be important, so the isobaric approximation is preferable. For geophysical applications, then, we choose the isobaric version:

$\rho C_{p} \frac{D T}{D t}=k \nabla^{2} T-\vec{\nabla} \cdot \vec{q}_{r a d}+\rho \varepsilon.\label{eqn:10}$

This is a generalization of the “heat equation’’ often discussed in physics texts: neglecting the radiation and dissipation terms, it becomes

$\frac{D T}{D t}=\kappa_{T} \nabla^{2} T.\label{eqn:11}$

Here, $$\kappa_T$$ is the thermal diffusivity, given by

$\kappa_{T}=\frac{k}{\rho C_{p}}=\left\{\begin{array}{ll} 1.4 \times 10^{-7} m^{2} s^{-1}, & \text {in water } \\ 1.9 \times 10^{-5} m^{2} s^{-1}, & \text {in air } \end{array}\right.\label{eqn:12}$

By using Equation $$\ref{eqn:10}$$ instead of $$\ref{eqn:1}$$ we are able to impose energy conservation while adding only one new unknown (instead of two), so at least we are no worse off in terms of closure. We now have 5 equations for 6 unknowns.

In the special case of incompressible fluid, the condition $$\vec{\nabla}\cdot\vec{u}=0$$ represents an additional equation and the set is closed (i.e., no more equations are needed). For the more general case, we need to make a further assumption about the nature of the fluid. This will take the form of an equation of state. Again, there is more than one possibility.